1 Knight Moves
Problem F- Knight Moves
题目来源:https://vjudge.net/contest/207868#problem/F
Problem description:
题意概括:中国象棋中的‘马’走日字格,从一个点到另一个点最短需要多少步。
解题思路:利用广度优先搜索解这道题
AC代码:
#include<stdio.h> #include<string.h> struct note { int x; int y; int s; }; int main(void) { struct note que[70]; char f, c; int b, d; int a[10][10], book[10][10]; int next[8][2] = {{-1, -2}, {-1, 2},{1,-2}, {1, 2}, {-2, -1}, {-2, 1}, {2, -1},{2, 1}}; int head, tail; int flag, i, tx, ty; while(scanf("%c%d %c%d", &f,&b, &c, &d)!= EOF) { memset(a, 0, sizeof(a)); memset(book, 0, sizeof(book)); head = 1; tail = 1; que[tail].x = b; que[tail].y = f - 'a' + 1; que[tail].s = 0; tail ++; book[b][f - 'a' + 1] = 1; flag = 0; while(head < tail) { if(f == c && b == d) { que[tail - 1].s = 0; break; } for(i = 0; i <= 7; i ++) { tx = que[head].x +next[i][0]; ty = que[head].y +next[i][1]; if(tx < 1 || tx >8 || ty < 1 || ty > 8) { continue; } if(book[tx][ty] == 0) { book[tx][ty] = 1; que[tail].x = tx; que[tail].y = ty; que[tail].s =que[head].s + 1; tail ++; } if(tx == d && ty+ 'a' - 1 == c) { flag = 1; break; } } if(flag == 1) { break; } head ++; } printf("To get from %c%d to%c%d takes %d knight moves.\n", f, b, c, d, que[tail - 1].s); getchar(); } return 0; }
错误原因:忘记了输入%c时要吃掉回车。